March 15th, 2016, 04:40 PM  #11  
Global Moderator Joined: Dec 2006 Posts: 16,788 Thanks: 1238  Quote:
 
March 15th, 2016, 04:46 PM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,555 Thanks: 2148 Math Focus: Mainly analysis and algebra 
The reason the article doesn't mention it is because the enumerated sequences are of infinite length. The list cannot be longer than that.

March 15th, 2016, 05:12 PM  #13 
Global Moderator Joined: Dec 2006 Posts: 16,788 Thanks: 1238  You don't have to prove it when the sequences and rows are both infinite. You must already be aware of that, as your previous post states "There is no number infinity. You can't treat infinity as if it were a finite number."

March 17th, 2016, 08:25 PM  #14  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 974 Thanks: 78  Quote:
If it is wider than long, Cantors argument doesn't work: post #7 Making statements about an infinite list of elements infinitely long is meaningless.  
March 17th, 2016, 08:45 PM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,555 Thanks: 2148 Math Focus: Mainly analysis and algebra 
Zylo. The existence of at least one infinite set is an axiom of set theory. You are welcome to deny that axiom, but if you do you are working in a different system to the one that Cantor was working in. If you deny the existence of infinite sets, Cantor is not wrong, he is simply irrelevant because if infinite sets do not exist, you obviously can't have a theorem about countably infinite sets of infinite series. If you don't deny the existence of infinite sets, then infinite lists and infinite sequences come with the basic axioms. Again, you can't complain about Cantor's use of them. What is meaningless is your persistent talking of rubbish about things you simply don't understand. 
March 17th, 2016, 11:32 PM  #16 
Global Moderator Joined: Dec 2006 Posts: 16,788 Thanks: 1238  That's not what you stated in post #4. You stated "Cantor's Diagonal Argument only works if the enumeration is wider than it is long." That assertion is untrue, as Cantor's diagonal argument works when the length of each sequence is a finite constant that equals the number of sequences. Also, the wikipedia article you gave as a reference for your false assertion doesn't make that assertion or any similar assertion. That's obviously untrue if the length and width are finite. Your justification doesn't make sense in that case. That observation is sufficient to show that your post #7 is wrong. You give no reason for this statement. It's "circular" anyway, so it's an invalid assertion by one of your own principles. 
March 17th, 2016, 11:58 PM  #17 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 974 Thanks: 78 
From post #4 (longer than wide) 1 0 1 0 1 0 1 0 0 1 1 1 0 0 1 . . . . s = 0 0 1 is in the list From post#7 Wider than long) 1) 0 1 0 0 1 1 0.............. .) 1 1 0 1 0 0 1.............. n) 0 0 1 0 1 0 1.............. n+1) 1 0 0 1 0 0 1.............. No matter what n is, Cantors sequence n+1 is in the list. Cantor's diagonal argument doesn't lead to a contradiction. 
March 18th, 2016, 12:10 AM  #18 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 974 Thanks: 78 
In ZF, the "axiom of infinity" defines infinity as an inductive set. https://en.wikipedia.org/wiki/Axiom_of_infinity 
March 18th, 2016, 12:27 AM  #19 
Global Moderator Joined: Dec 2006 Posts: 16,788 Thanks: 1238  
March 18th, 2016, 01:19 AM  #20 
Global Moderator Joined: Dec 2006 Posts: 16,788 Thanks: 1238  You given no reason for that assertion, whereas Cantor gives a straightforward reason why it isn't  its "diagonal" construction ensures it differs from the first sequence in the first place, the second sequence in the second place, the nth sequence in the nth place, the n+1th sequence in the n+1th place, etc., so that it can't match any sequence already listed.


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argument, cantor, diagonal, infinity, number 
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